ÌâÄ¿ÄÚÈÝ

СÀèͬѧΪÁ˽øÒ»²½¼ÓÉî¶Ô¡°¼îµÄ»¯Ñ§ÐÔÖÊ¡±µÄÀí½â£¬ÌØÑûÄãЭÖúÍê³ÉÏÂÁлÓë̽¾¿£º

£¨1£© ÈçÓÒͼËùʾ£¬ÔÚ°×É«µãµÎ°åÉϽøÐÐʵÑ飬Ç뽫ʵÑéÏÖÏóÌîÈëÏÂ±í£º

£¨2£©»ØÒä¼ìÑé¶þÑõ»¯Ì¼ÆøÌåµÄ·´Ó¦£¬Ð´³ö¸Ã·´Ó¦µÄ»¯Ñ§·½³Ìʽ                          ¡£

£¨3£©ÈýÑõ»¯Áò£¨SO3£©ÓëÇâÑõ»¯ÄÆ·´Ó¦ÓëÉÏÃæµÄ·´Ó¦ÀàËÆ£¬Ð´³öÕâÒ»·´Ó¦µÄ»¯Ñ§·½³Ìʽ?                                   ¡£?

£¨4£©ÈçÓÒͼËùʾ£¬ÔÚÉÕ±­ÖмÓÈë10mLÇâÑõ»¯ÄÆÈÜÒº£¬µÎÈ뼸µÎ·Ó̪ÊÔÒº£¬ÈÜÒºÏÔ     É«£¬ÔÙÓõιÜÂýÂýµÎÈëÏ¡ÑÎËᣬ²¢²»¶Ï½Á°èÈÜÒº£¬ÖÁÈÜÒºÑÕÉ«Ç¡ºÃ±ä³É

       ÎªÖ¹¡£

ÕâһʵÑé˵Ã÷£ºËáÓë¼î×÷ÓÃÉú³ÉÁËÑκÍË®£¬ÕâÒ»·´Ó¦½Ð×ö        ·´Ó¦¡£?Öкͷ´Ó¦µÄʵÖÊÊÇ                             £»ÊµÑéÖеμӷÓ̪µÄ×÷ÓÃÊÇ                     ¡£ 

£¨5£©¸ù¾ÝÉÏÃæµÄʵÑéºÍÌÖÂÛ£¬ÊÔ¹éÄɳöÇâÑõ»¯ÄÆ¡¢ÇâÑõ»¯¸ÆÓÐÄÄЩÏàËƵĻ¯Ñ§ÐÔÖÊ¡££¨ÈÎдÁ½µã£©?

¢Ù                                       ¢Ú                        

 

(1)˦ ˦

£¨2£©Ca(OH)2+CO2=CaCO3+H2O

£¨3£©SO3+2NaOH=Na2SO4+H2O

£¨4£©ºì ÖÐºÍ ÇâÀë×ÓºÍÇâÑõ¸ùÀë×Ó·´Ó¦Éú³ÉË®

£¨5£©ÄܺÍËáÐÔÆøÌå·¢Éú·´Ó¦£»ÄܺÍËá·¢ÉúÖкͷ´Ó¦

½âÎö:£¨1£©¸ù¾ÝʯÈïÊÔÒºÓöËáÐÔÈÜÒº±äºì£¬Óö¼îÐÔÈÜÒº±äÀ¶£¬ÓöÖÐÐÔÈÜÒº²»±äÉ«Åж¨£¬£¨2£©Êéд»¯Ñ§Ê½Ó¦¸Ã×¢Ò⻯ѧʽ¡¢Åäƽ¡¢Ìõ¼þ¡¢¼ýÍ·£¬

£¨3£©·Ó̪ÊÔÒºÓöËáÐÔÈÜÒº²»±äÉ«£¬Óö¼îÐÔÈÜÒº±äºì£¬ÓöÖÐÐÔÈÜÒº²»±äÉ«£®Öкͷ´Ó¦ÊÇÖ¸ËáÓë¼îÉú³ÉÑκÍË®µÄ·´Ó¦£¬ÊµÖÊÊÇÇâÀë×ÓºÍÇâÑõ¸ùÀë×Ó·´Ó¦Éú³ÉË®¡£

£¨4£©ÇâÑõ»¯ÄƺÍÇâÑõ»¯¸Æ¶¼ÊôÓڼ¾ßÓмîµÄͨÐÔ£º¢ÙʹËá¼îָʾ¼ÁÏÔʾ²»Í¬µÄÑÕÉ«£»¢ÚÓëijЩ·Ç½ðÊôÑõ»¯Îï·´Ó¦£»  ¢ÛÓëËá·¢ÉúÖкͷ´Ó¦£»¢ÜÓëijЩÑη¢Éú¸´·Ö½â·´Ó¦

½øÐзÖÎö½â´ð

 

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
£¨2009?¦µ×£©Ð¡ÀèͬѧΪÁ˽øÒ»²½¼ÓÉî¶Ô¡°¼îµÄ»¯Ñ§ÐÔÖÊ¡±µÄÀí½â£¬ÌØÑûÄãЭÖúÍê³ÉÏÂÁлÓë̽¾¿£º?
£¨1£©ÈçͼËùʾ£¬ÔÚ°×É«µãµÎ°åÉϽøÐÐʵÑ飬Ç뽫ʵÑéÏÖÏóÌîÈëÏÂ±í£º
ÇâÑõ»¯ÄÆÈÜÒºÇâÑõ»¯¸ÆÈÜÒº
¼Ó×ÏɫʯÈïÈÜÒº
ÈÜÒº±äÀ¶
ÈÜÒº±äÀ¶
ÈÜÒº±äÀ¶
ÈÜÒº±äÀ¶
£¨2£©»ØÒä¼ìÑé¶þÑõ»¯Ì¼ÆøÌåµÄ·´Ó¦£¬Ð´³ö¸Ã·´Ó¦µÄ»¯Ñ§·½  ³Ìʽ
CO2+Ca£¨OH£©2=CaCO3¡ý+H2O
CO2+Ca£¨OH£©2=CaCO3¡ý+H2O
£®?
£¨3£©ÈýÑõ»¯Áò£¨SO3£©ÓëÇâÑõ»¯ÄÆ·´Ó¦ÓëÉÏÃæµÄ·´Ó¦ÀàËÆ£¬Ð´³öÕâÒ»·´Ó¦µÄ»¯Ñ§·½³Ìʽ
SO3+2NaOH=Na2SO4+H2O
SO3+2NaOH=Na2SO4+H2O
£®
£¨4£©ÈçͼËùʾ£¬ÔÚÉÕ±­ÖмÓÈë10mLÇâÑõ»¯ÄÆÈÜÒº£¬µÎÈ뼸µÎ·Ó̪ÊÔÒº£¬ÈÜÒºÏÔ
ºì
ºì
É«£¬ÔÙÓõιÜÂýÂýµÎÈëÏ¡ÑÎËᣬ²¢²»¶Ï½Á°èÈÜÒº£¬ÖÁÈÜÒºÑÕÉ«Ç¡ºÃ±ä³ÉÎÞɫΪֹ£®ÕâһʵÑé˵Ã÷£ºËáÓë¼î×÷ÓÃÉú³ÉÁËÑκÍË®£¬ÕâÒ»·´Ó¦½Ð×ö
ÖкÍ
ÖкÍ
·´Ó¦£®
£¨5£©¸ù¾ÝÉÏÃæµÄʵÑéºÍÌÖÂÛ£¬ÊÔ¹éÄɳöÇâÑõ»¯ÄÆ¡¢ÇâÑõ»¯¸ÆÓÐÄÄЩÏàËƵĻ¯Ñ§ÐÔÖÊ£®£¨ÈÎдÁ½µã£©?
¢Ù
¶¼ÄÜʹָʾ¼Á±äÉ«
¶¼ÄÜʹָʾ¼Á±äÉ«

¢Ú
¶¼ÄܺÍijЩ·Ç½ðÊôÑõ»¯Îï·´Ó¦
¶¼ÄܺÍijЩ·Ç½ðÊôÑõ»¯Îï·´Ó¦
£®

СÀèͬѧΪÁ˽øÒ»²½¼ÓÉî¶Ô¡°¼îµÄ»¯Ñ§ÐÔÖÊ¡±µÄÀí½â£¬ÌØÑûÄãЭÖúÍê³ÉÏÂÁлÓë̽¾¿£º?

£¨1£© ÈçÓÒͼËùʾ£¬ÔÚ°×É«µãµÎ°åÉϽøÐÐʵÑ飬Ç뽫ʵÑéÏÖÏóÌîÈëÏÂ±í£º

         

ÇâÑõ»¯ÄÆÈÜÒº

ÇâÑõ»¯¸ÆÈÜÒº

¼Ó×ÏɫʯÈïÈÜÒº

            

           

£¨2£©»ØÒä¼ìÑé¶þÑõ»¯Ì¼ÆøÌåµÄ·´Ó¦£¬Ð´³ö¸Ã·´Ó¦µÄ»¯Ñ§·½³Ìʽ                                              ¡£?

£¨3£©ÈýÑõ»¯Áò£¨SO3£©ÓëÇâÑõ»¯ÄÆ·´Ó¦ÓëÉÏÃæµÄ·´Ó¦ÀàËÆ£¬Ð´³öÕâÒ»·´Ó¦µÄ»¯Ñ§·½³Ìʽ?                                          ¡£?

£¨4£©ÈçÓÒͼËùʾ£¬ÔÚÉÕ±­ÖмÓÈë10mLÇâÑõ»¯ÄÆÈÜÒº£¬µÎÈ뼸µÎ·Ó̪ÊÔÒº£¬ÈÜÒºÏÔ     É«£¬ÔÙÓõιÜÂýÂýµÎÈëÏ¡ÑÎËᣬ²¢²»¶Ï½Á°èÈÜÒº£¬ÖÁÈÜÒºÑÕÉ«Ç¡ºÃ±ä³ÉÎÞɫΪֹ¡£ÕâһʵÑé˵Ã÷£ºËáÓë¼î×÷ÓÃÉú³ÉÁËÑκÍË®£¬ÕâÒ»·´Ó¦½Ð×ö      ·´Ó¦¡£

£¨5£©¸ù¾ÝÉÏÃæµÄʵÑéºÍÌÖÂÛ£¬ÊÔ¹éÄɳöÇâÑõ»¯ÄÆ¡¢ÇâÑõ»¯¸ÆÓÐÄÄЩÏàËƵĻ¯Ñ§ÐÔÖÊ¡££¨ÈÎдÁ½µã£©

¢Ù                                                

¢Ú                                                

 

СÀèͬѧΪÁ˽øÒ»²½¼ÓÉî¶Ô¡°¼îµÄ»¯Ñ§ÐÔÖÊ¡±µÄÀí½â£¬ÌØÑûÄãЭÖúÍê³ÉÏÂÁлÓë̽¾¿£º

£¨1£© ÈçÓÒͼËùʾ£¬ÔÚ°×É«µãµÎ°åÉϽøÐÐʵÑ飬Ç뽫ʵÑéÏÖÏóÌîÈëÏÂ±í£º

         

ÇâÑõ»¯ÄÆÈÜÒº

ÇâÑõ»¯¸ÆÈÜÒº

¼Ó×ÏɫʯÈïÈÜÒº

            

           

£¨2£©»ØÒä¼ìÑé¶þÑõ»¯Ì¼ÆøÌåµÄ·´Ó¦£¬Ð´³ö¸Ã·´Ó¦µÄ»¯Ñ§·½³Ìʽ                                              ¡£

£¨3£©ÈýÑõ»¯Áò£¨SO3£©ÓëÇâÑõ»¯ÄÆ·´Ó¦ÓëÉÏÃæµÄ·´Ó¦ÀàËÆ£¬Ð´³öÕâÒ»·´Ó¦µÄ»¯Ñ§·½³Ìʽ                                           ¡£

£¨4£©ÈçÓÒͼËùʾ£¬ÔÚÉÕ±­ÖмÓÈë10mLÇâÑõ»¯ÄÆÈÜÒº£¬µÎÈ뼸µÎ·Ó̪ÊÔÒº£¬ÈÜÒºÏÔ      É«£¬ÔÙÓõιÜÂýÂýµÎÈëÏ¡ÑÎËᣬ²¢²»¶Ï½Á°èÈÜÒº£¬ÖÁÈÜÒºÑÕÉ«Ç¡ºÃ±ä³ÉÎÞɫΪֹ¡£ÕâһʵÑé˵Ã÷£ºËáÓë¼î×÷ÓÃÉú³ÉÁËÑκÍË®£¬ÕâÒ»·´Ó¦½Ð×ö      ·´Ó¦¡£

£¨5£©¸ù¾ÝÉÏÃæµÄʵÑéºÍÌÖÂÛ£¬ÊÔ¹éÄɳöÇâÑõ»¯ÄÆ¡¢ÇâÑõ»¯¸ÆÓÐÄÄЩÏàËƵĻ¯Ñ§ÐÔÖÊ¡££¨ÈÎдÁ½µã£©

¢Ù                                                 

¢Ú                                                 

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø